\newproblem{lay:5_4_5}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.4.5}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $T:\mathbb{P}_2\rightarrow\mathbb{P}_3$ be the transformation that maps a polynomial $p(t)$ into the polynomial $(t+3)p(t)$.
	\begin{enumerate}[a.]
		\item Find the image of $p(t)=3-2t+t^2$
		\item Show that $T$ is a linear transformation
		\item Find the matrix for $T$ relative to the bases $\{1,t,t^2\}$ and $\{1,t,t^2,t^3\}$.
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}[a.]
		\item $T(3-2t+t^2)=(t+3)(3-2t+t^2)=9-3t+t^2+t^3$
		\item We need to show that $T(p_1(t)+p_2(t))=T(p_1(t))+T(p_2(t))$ and $T(c(p(t))=cT(p(t))$
			    \begin{itemize}
						\item $T(p_1(t)+p_2(t))=T(p_1(t))+T(p_2(t))$
							\begin{center}
								$\begin{array}{rcl}
									T(p_1(t)+p_2(t))&=&(t+3)(p_1(t)+p_2(t))=(t+3)p_1(t)+(t+3)p_2(t)\\
									   &=&T(p_1(t))+T(p_2(t))\\
								\end{array}$
							\end{center}
						\item $T(c(p(t))=cT(p(t))$
							\begin{center}
								$\begin{array}{rcl}
									T(c(p(t))&=&(t+3)(cp(t))=c(t+3)p(t)=c((t+3)p(t))=cT(p(t))\\
								\end{array}$
							\end{center}
						\end{itemize}
			\item We need to calculate the transformation of each of the elements in the basis $\{1,t,t^2\}$
							\begin{center}
								$\begin{array}{lcl}
									T(1)&=&(t+3)1=t+3 \\
									T(t)&=&(t+3)t=t^2+3t \\
									T(t^2)&=&(t+3)t^2=t^3+3t^2 \\
								\end{array}$
							\end{center}
							The matrix sought is
							\begin{center}
								$M=\begin{pmatrix} 3 & 0 & 0 \\ 1 & 3 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{pmatrix}$
							\end{center}
	\end{enumerate}
}
\useproblem{lay:5_4_5}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
